Find the first 3 terms in the expansion of (2x - x)° in ascending powers of x. (ii) Hence find the coefficient of in the expansion of (2 + x)(2x-x).
(2x-x)° = 1 for all x≠0
Maybe you can clarify ...
To find the first 3 terms in the expansion of (2x - x)° in ascending powers of x, we use the binomial theorem.
The binomial theorem states that for any two numbers a and b, and any positive integer n, the expansion of (a + b)ⁿ can be written as:
(a + b)ⁿ = C(n, 0) * aⁿ * b⁰ + C(n, 1) * aⁿ⁻¹ * b¹ + C(n, 2) * aⁿ⁻² * b² + ...
In this case, a = 2x and b = -x, so the expansion we want to find is (2x - x)⁰.
Using the binomial theorem, we can write the expansion as:
(2x - x)⁰ = C(0, 0) * (2x)⁰ * (-x)⁰
C(0, 0) is the binomial coefficient, which is equal to 1 when the top number is 0.
So the first term in the expansion is 1 * (2x)⁰ * (-x)⁰ = 1.
To find the next term, we substitute n = 1 into the binomial theorem formula:
(2x - x)¹ = C(1, 0) * (2x)¹ * (-x)⁰ + C(1, 1) * (2x)⁰ * (-x)¹
C(1, 0) = 1 and C(1, 1) = 1, so the second term is 1 * (2x)¹ * (-x)⁰ + 1 * (2x)⁰ * (-x)¹ = 2x - x.
To find the third term, we substitute n = 2 into the binomial theorem formula:
(2x - x)² = C(2, 0) * (2x)² * (-x)⁰ + C(2, 1) * (2x)¹ * (-x)¹ + C(2, 2) * (2x)⁰ * (-x)²
C(2, 0) = 1, C(2, 1) = 2, and C(2, 2) = 1, so the third term is 1 * (2x)² * (-x)⁰ + 2 * (2x)¹ * (-x)¹ + 1 * (2x)⁰ * (-x)² = 4x² - 4x² + x² = x².
Therefore, the first 3 terms in the expansion of (2x - x)° in ascending powers of x are 1, 2x - x, x².
(i) Now, let's find the coefficient of x² in the expansion of (2 + x)(2x - x).
To do this, we can multiply the two expressions together:
(2 + x)(2x - x) = 2 * 2x - 2 * x + x * 2x - x * x = 4x - 2x + 2x² - x² = 2x² + 2x - x²
The coefficient of x² is the number in front of x², which is 2 - 1 = 1.
Therefore, the coefficient of x² in the expansion of (2 + x)(2x - x) is 1.
To find the first 3 terms in the expansion of (2x - x)^0 in ascending powers of x, we can simplify the expression (2x - x)^0 to 1. Therefore, the first term is 1.
Now, let's find the second term. The second term in the expansion of (2x - x)^0 can be found using the binomial theorem. According to the binomial theorem, the second term is given by the expression (0C1) * (2x)^0 * (-x)^1, where 0C1 is the binomial coefficient of choosing 1 term from 0 terms.
Simplifying the expression, we have:
Second term = (0C1) * (2x)^0 * (-x)^1
= 0 * 1 * (-x)
= 0
Therefore, the second term is 0.
To find the third term, we follow the same process. The formula to find the term is (0C2) * (2x)^0 * (-x)^2, where 0C2 is the binomial coefficient of choosing 2 terms from 0 terms.
Simplifying the expression, we have:
Third term = (0C2) * (2x)^0 * (-x)^2
= 0 * 1 * x^2
= 0
Therefore, the third term is 0.
Now, let's find the coefficient of x^2 in the expansion of (2 + x)(2x - x). We multiply the expressions (2 + x) and (2x - x) using the distributive property.
(2 + x)(2x - x) = 2(2x - x) + x(2x - x)
Expanding this expression, we have:
(2 + x)(2x - x) = 4x - 2x^2 + 2x^2 - x^2
The terms involving x^2 cancel out, leaving:
(2 + x)(2x - x) = 4x
Therefore, the coefficient of x^2 in the expansion of (2 + x)(2x - x) is 0.